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2 years ago

Someone plz give me some boy advice I really need it.

1 answer:

8 0

If your wanting to know how to talk to them, just act like you know them really well be very causal and be confident in what you say, they also like it when you make the first move to talk to them

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A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th

Alika [10]

Answer:

Wave A

I hope this helps

8 0

2 years ago

Estimate how long a 2500 W electric kettle would take to boil away 1.5 Kg of water . The specific latent heat of vaporization of

andrew-mc [135]

The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

<h3>How to calculate the time</h3>

Use the formula:

Power × time = mass × specific heat

Given mass = 1. 5kg

Specific latent heat of vaporization = 4000000 J/ Kg

Power = 2500 W

Substitute the values into the formula

Power × time = mass × specific heat

2500 × time = 1. 5 × 4000000

Make 'time' the subject

time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds

Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

Learn more about specific latent heat of vaporization:

https://brainly.in/question/1580957

#SPJ1

8 0

2 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

2. In the activities below indicate if friction is useful or not useful? Explain your answer. i. Writing yes it is useful becaus

Gemiola [76]

Answer:

Please find the answer in the explanation

Explanation:

Friction is a force that opposes motion. One or two of the advantages of friction are break and ability of an object to walk.

Writing yes it is useful because when your writing because friction helps you see what your writing

ii. Rubbing. Yes, it is useful.

friction make it possible for two object to rub each other

iii. Skiing. No. It is not useful because With presence of friction, skiing will not be possible.

iv. Rotating a wheel No. It is not useful because Friction will oppose the rotation of the wheel.

8 0

3 years ago

The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m

butalik [34]

$\sf\underline{Solution:}$

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

$\sf{Now,according \: to \:the\:question:}$

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

☆ Substitute the required values in the given formula-

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0

2 years ago